So basically I'd like to sum two numbers and return their value while using a void function in C. I know this is easy peasy by using a normal function returning an int or a numeric type but I wanna work on my pointer knowledge.
I tried creating a pointer inside main() and then passing it as an argument to the void function. Then I calculated my sum in a new int variable and assigned the pointer to point to that specific variable. The problem is I can't "retrieve" it or "find" that area of memory in the main function.
Here's what I've tried:
void testFunction(int a,int b, int *x)
{
int c=a+b;
x=&c;
}
int main()
{
int n1=7;
int n2=90;
int *pointerParam;
testFunction(n1, n2, pointerParam);
printf("Value of pointer is %d\n", *pointerParam);
}
It just exits with an error code, it does nothing. If I try to printf *x inside the function, it does work so I know that part at least works.
Any help would be greatly appreciated!
There are multiple problems with the code as it is shown.
The main problem is probably that you misunderstand how emulation of pass-by-reference works in C.
For it to work you need to pass a pointer to the variable that should be set. This is done using the pointer-to operator &. You also need to dereference the pointer, to set the variable the pointer is pointing to.
Putting it together your program should look something like this (simplified):
void testFunction(int a,int b, int *x)
{
// Assign to where `x` is pointing
*x = a + b;
}
int main(void)
{
int n1 = 7;
int n2 = 90;
int result; // Where the result should be written
// Pass a pointer to the `result` variable, so the function can write to it
testFunction(n1, n2, &result);
}
I will show you a program that is no more than a set of printf() and your function. Maybe the program's output helps in showing these pointers, arrays and integers things.
The example
Consider these variables
int n[4] = {10, 20, -30, -40};
int v1 = 0;
int v2 = 0;
int* a_pointer = NULL;
int* is a pointer to an int, that's the meaning of the asterisk in the declaration. In this context the asterisk is called the dereference operator or the indirection operator. But the asterisk also is the multiplication operator in C. ;)
Now a_pointer points to nothing, the meaning of the NULL.
But a_pointer is there to hold an address of something, of an int. The way of getting such address of something is the address of operator, the &, that also has an alternate life as the bitwise and operator in C. Things of life.
In printf() the %p specifier shows an address. This
printf("\naddress of v1 is %p\n", &v1);
printf("address of v2 is %p\n", &v2);
printf("address of array n[0] is %p\n", n);
printf("address of array n[0] is %p\n", 1 + n);
printf("address of array n[0] is %p\n", 2 + n);
printf("address of array n[0] is %p\n\n", 3 + n);
shows (in a 32-bits compilation)
address of v1 is 008FFDA0
address of v2 is 008FFD9C
address of array n[0] is 008FFDA4
address of array n[1] is 008FFDA8
address of array n[2] is 008FFDAC
address of array n[3] is 008FFDB0
And you will see the reason the program prints these lines in the code below...
This line
a_pointer = &v1;
takes the address of v1 and assign it to the pointer a_pointer.
Now a_pointer is pointing to something, to v1, and you can use it in your function. These lines are equivalent
testFunction(n[0], n[3], &v1);
and
testFunction(n[0], n[3], a_pointer);
In these lines
testFunction(n[0], n[3], &v1);
printf("n[0] = %d, n[3] = %d, sum in v1 is %d\n", n[0], n[3], v1);
printf(
"p points to v1. value is %4d, address is %p\n\n", *a_pointer,
a_pointer);
you see the use of the pointer to access the value it points to, using the dereference operator in the printf().
Follow the program along to see a few uses of this.
In particular, see these lines
a_pointer = n + 3;
printf(
"\np points now to n[3]. value is %4d, address is %p\n",
*a_pointer, a_pointer);
to see the thing C is made for: address memory easily. n is int[4], an array of int. a_pointer is a pointer to int. And the language knows that when you write a_pointer = n + 3 that it needs to add to the address of n, an int[], the size of 3 int variables, and assign it to the pointer, so *a_pointer is n[3]and it is used to call testFunction() in
testFunction(n[1], n[2], n + 3);
program output
n[] is [10,20,-30,-40], v1 is 0 v2 is 0
address of v1 is 00CFFA9C
address of v2 is 00CFFA98
address of array n[0] is 00CFFAA0
address of array n[1] is 00CFFAA4
address of array n[2] is 00CFFAA8
address of array n[3] is 00CFFAAC
n[0] = 10, n[3] = -40, sum in v1 is -30
p points to v1. value is -30, address is 00CFFA9C
p now points to v2. value is 0, address is 00CFFA98
n[0] = 10, n[1] = 20, sum in v2 is 30
n[] is [10,20,-30,-40], v1 is -30 v2 is 30
p now points to v1. value is -30, address is 00CFFA9C
n[] is [10,20,-30,-40], v1 is -30 v2 is 30
now makes n[3] = n[1] + n[2] using testFunction()
n[] is [10,20,-30,-10], v1 is -30 v2 is 30
p points now to n[3]. value is -10, address is 00CFFAAC
the code
#include <stdio.h>
show(int[4], int, int);
void testFunction(int, int, int*);
int main(void)
{
int n[4] = {10, 20, -30, -40};
int v1 = 0;
int v2 = 0;
int* a_pointer = NULL;
show(n, v1, v2);
printf("\naddress of v1 is %p\n", &v1);
printf("address of v2 is %p\n", &v2);
printf("address of array n[0] is %p\n", n);
printf("address of array n[1] is %p\n", 1 + n);
printf("address of array n[2] is %p\n", 2 + n);
printf("address of array n[3] is %p\n\n", 3 + n);
a_pointer = &v1;
testFunction(n[0], n[3], &v1);
printf("n[0] = %d, n[3] = %d, sum in v1 is %d\n", n[0], n[3], v1);
printf(
"p points to v1. value is %4d, address is %p\n\n", *a_pointer,
a_pointer);
a_pointer = &v2;
printf(
"p now points to v2. value is %4d, address is %p\n", *a_pointer,
a_pointer);
testFunction(n[0], n[1], &v2);
printf("n[0] = %d, n[1] = %d, sum in v2 is %d\n", n[0], n[1], v2);
show(n, v1, v2);
a_pointer = &v1;
printf(
"\np now points to v1. value is %4d, address is %p\n", *a_pointer,
a_pointer);
show(n, v1, v2);
printf("\nnow makes n[3] = n[1] + n[2] using testFunction()\n");
testFunction(n[1], n[2], n + 3);
show(n, v1, v2);
a_pointer = n + 3;
printf(
"\np points now to n[3]. value is %4d, address is %p\n",
*a_pointer, a_pointer);
return 0;
};
show(int n[4], int v1, int v2)
{
printf(
"n[] is [%d,%d,%d,%d], v1 is %d v2 is %d\n", n[0], n[1], n[2],
n[3], v1, v2);
};
void testFunction(int a, int b, int* sum)
{
*sum = a + b;
return;
}
I will not go into religious discussions here, but you may find easier to understand the meaning of the declarions if you write
int* some_int = NULL;
instead of
int *some_int = NULL;
you declare a name, and the name is some_int. The compiler will tell you that some_int is int*, its type. The fact that *some_int is an int is a consequence of the application of an operator to a variable.
You lacked just to understand how pointers are managed, but you were almost correct:
/* this is almost correct, you could have just said: *x = a + b; */
void testFunction(int a,int b, int *x)
{
int c=a+b;
*x=c; /* the pointed to value is what we are assigning */
}
int main()
{
int n1=7;
int n2=90;
int result; /* vvvvvvv this is the important point */
testFunction(n1, n2, &result); /* you pass the address of result as the required pointer */
printf("Value of pointer is %d\n", result);
}
Trying to double dereference and print them (TOP TWO ARE EXAMPLES):
printf ("a's value = %d \n", a) ;
printf ("a's address = %p \n", &a) ;
printf ("a_ptr_ptr deref'ed defer'ed =d% \n",
What would go after the \n", for a_ptr_ptr deref'ed defer'ed
If you want the address of the address of a, you're going to have to store a's address in a pointer variable, and take the address of that. But having done so, yes, you can double-dereference that pointer with **, and get a's value back. Something like this:
int a = 5;
int *ip = &a;
int **ipp = &ip;
printf("ipp = %p\n", ipp);
printf("*ipp = %p, ip = %p, &a = %p\n", *ipp, ip, &a);
printf("**ipp = %d, *ip = %d\n", **ipp, *ip);
Theoretically you can continue this as long as you like:
int ***ippp = &ipp;
int ****ipppp = &ippp;
int *****ippppp = &ipppp;
printf("*****ippppp = %d\n", *****ippppp);
But by now this is mostly a game; there's no practical use in a real C program for a 5-level pointer, and at some point (after 8 or 10 levels, I think) the compiler's allowed to say "All right, enough, game over!".
I want to assign the struct a to the struct b. Printing the address and the values of the array before and after function call it shows that during the function call the assignment works and both pointers point to the same struct address. However, after returning from the function the changes are reversed. Why?
typedef struct arrayA {
int a[3];
}arrayA;
void f(arrayA *a, arrayA *b){
a = b;
printf("address of a: %p\n", a);
printf("address of b: %p\n", b);
}
int main(int argc, char *argv[]) {
printf("------------ Array assignment test -----------------\n");
arrayA a = { { 0, 0, 0} };
arrayA b = { { 1, 1, 1} };
printf("address of a: %p\n", &a);
printf("address of b: %p\n", &b);
printf("a[0] : %d a[1] : %d\n", a.a[0], a.a[1]);
f(&a, &b);
printf("a[0] : %d a[1] : %d\n", a.a[0], a.a[1]);
printf("address of a: %p\n", &a);
printf("address of b: %p\n", &b);
printf("----------------------------------------------------\n");
return 0;
}
Prints
------------ Array assignment test -----------------
address of a: 0x7ffd3fc17b80
address of b: 0x7ffd3fc17b90
a[0] : 0 a[1] : 0
address of a: 0x7ffd3fc17b90
address of b: 0x7ffd3fc17b90
a[0] : 0 a[1] : 0
address of a: 0x7ffd3fc17b80
address of b: 0x7ffd3fc17b90
----------------------------------------------------
You are passing the pointers by value and expecting them to get modified. They will be modified within the function as there is a copy of the pointer (address) that is made local to the function. When the function returns the originals remain unchanged. (You could try to print the address of the local variables within the function to understand better.)
If you want to change the structs, you'll want to dereference the pointers:
void f(arrayA *a, arrayA *b){
*a = *b;
printf("address of a: %p\n", a);
printf("address of b: %p\n", b);
}
If you want to change the pointers themselves, you'll need an extra indirection:
void f(arrayA **a, arrayA **b){ // Note the ** here
*a = *b;
printf("address of a: %p\n", a);
printf("address of b: %p\n", b);
}
However, after returning from the function the changes are reversed. Why?
Your f() function really just change the pointer inside it, and those pointer changes are not reserved after the function.
You can copy struct by pass-by-pointer:
void f(arrayA *a, arrayA *b){
*a = *b;
}
That will ensure you can copy struct in main() via
f(&a, &b);
As all you need is copy struct, there is no need to print out the addresses at all. In case you do need to debug the address, you should convert to (void *) to avoid all the warnings with printf("%p")
Firstly, I've already reviewed these:
How are multi-dimensional arrays formatted in memory?
Memory map for a 2D array in C
From there, it is known that a 2D array is not the same as char** but in memory they look exactly same. This sounds strange and so I've researched the following:
#include <stdio.h>
char func(char** m) {
return m[0][0]; //only works for char**, already discussed in the other SO question
}
int main() {
//char a[4][2]; //a 2D char array
int row = 4, col = 2; //char**
char** a = malloc(row * sizeof(char*));
int i;
for (i = 0; i < row; i++) {
a[i] = malloc(col * sizeof(char));
}
//checking the output
printf(" &a = %u\n", &a);
printf(" &a[0] = %u\n", &a[0]);
printf("&a[0][0] = %u\n", &a[0][0]);
printf(" a = %u\n", a);
printf(" a[0] = %u\n", a[0]);
//printf(" a[0][0] = %u\n", a[0][0]); //char value in a[0][0], here a garbage value
//char m = func(a); //only works for char**, already discussed in the other SO question
return 0;
}
Possible output for char** :
&a = 3209288 // &a
&a[0] = 4083720 // &(*(a+0)) = a
&a[0][0] = 4083784 // &(*(*(a+0)+0)) = *a
a = 4083720 // a
a[0] = 4083784 // *(a+0) = *a
Possible output for 2D char array :
&a = 3473104 // &a
&a[0] = 3473104 // a
&a[0][0] = 3473104 // *a
a = 3473104 // a
a[0] = 3473104 // *a
It is easy to understand the output of char**. But the output of 2D char array looks strange though it was discussed in the other SO question. I cannot think of a pointer x of whatever data-type when,
x = &x = *x
and all the 3 things physically reside in the same block of memory. Hope that my confusion is understandable. Can anyone explain the mystery?
When you use name of the array is any expression except &array and sizeof array this name will be automaticly converted to pointer to the first element of the array (char [5][10] will be converted to char (*)[10]). Address of this pointer will be equal to the adress of entire array.
So, "sizeof (char [5][10]) == 50", there is no additional pointers.
char arr[5][10];
&a = 3473104 // Address of entire array, (char (*)[5][10])
&a[0] = 3473104 // Address of first row, (char (*)[10])
&a[0][0] = 3473104 // Address of first char, (char *)
a = 3473104 // "a", (char [5][10]), converted to "&a[0]", (char (*)[10])
a[0] = 3473104 // "a[0]", (char[10]), converted ro "&a[0][0]", (char *)
I put this code into eclipse and run it
main()
{
int *p, *q, *r;
int a = 10, b = 25;
int c[4] = {6,12,18,24};
p = c;
printf("p = %d\n" ,p);
}
the output I get is
p = 2358752
what is this number supposed to represent? Is it the address of the variable?
If what i'm saying above is true would my answer to the following question be correct?
so lets say the following are stored at the following locations
address variables
5000 p
5004 q
5008 r
500C a
5010 b
5014 c[0]
5018 c[1]
501C c[2]
5020 c[3]
so would would the line
p = c;
be 5014?
int *p,
The above statement defines p to be a pointer to an integer.
In the below statement, c is implicitly converted to a pointer to the first element of the array a.
p = c;
// equivalent to
p = &c[0];
Therefore, p contains the address of the first element of the array. Also, the conversion specifier to print an address is %p.
printf("p = %p\n", (void *)p);
// prints the same address
printf("c = %p\n", (void *)c);
Yes, p is the address of c, which is the same as the address of c[0]. And yes, in your second example, p would be equal to 5014.